How do you add $(8 + 8 i) + (- 4 + 6 i)$ $(8+8i)+(-4+6i)$ in trigonometric form?

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How do you add $(8 + 8 i) + (- 4 + 6 i)$ $(8+8i)+(-4+6i)$ in trigonometric form?

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Answer 1 · 2017-10-07T15:19:27

See below.

Explanation:

To convert complex numbers to trigonometric form, find $r$ $r$ , the distance of the point away from the origin, and $θ$ $theta$ , the angle.

$(8 + 8 i) + (- 4 + 6 i) = 8 - 4 + 8 i + 6 i = 4 + 14 i$ $(8 + 8i) + (-4 + 6i) = 8 - 4 + 8i + 6i = 4 + 14i$

$4 + 14 i$ $4 + 14i$ is in the form $a + b i$ $a+bi$ . First, find $r$ $r$ :

$r^{2} = a^{2} + b^{2}$ $r^2 = a^2 + b^2$

$r^{2} = 4^{2} + 14^{2}$ $r^2 = 4^2 + 14^2$

$r = \sqrt{252} = 2 \sqrt{53}$ $r = sqrt252 = 2sqrt53$

Find $θ$ $theta$ :

$tan θ = \frac{b}{a}$ $tan theta = b/a$

$tan θ = \frac{14}{4}$ $tan theta = 14/4$

$θ = {tan}^{- 1} (\frac{14}{4}) \approx 1.29$ $theta = tan^-1(14/4) ~~ 1.29$

In trigonometric form, this is $r (cos θ + i sin θ)$ $r(cos theta + i sin theta)$ or in shorthand,
$r$ $r$ $c i s$ $cis$ $θ$ $theta$ .

Thus the answer is $2 \sqrt{53} (cos 1.29 + i sin 1.29)$ $2sqrt53 (cos 1.29 + i sin 1.29)$ or $2 \sqrt{53}$ $2sqrt 53$ $c i s$ $cis$ $1.29$ $1.29$ .

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How do you add $(8 + 8 i) + (- 4 + 6 i)$ $(8+8i)+(-4+6i)$ in trigonometric form?

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Explanation:

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How do you add $(8 + 8 i) + (- 4 + 6 i)$ $(8+8i)+(-4+6i)$ in trigonometric form?